If n=110 and widehat{p} (p-hat) =0.66, construct a 95% confidence interval. Give your answers to two decimals ?

Question
Confidence intervals
asked 2020-10-25
If \(\displaystyle{n}={110}\) and \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}\) (p-hat) \(\displaystyle={0.66}\), construct a \(\displaystyle{95}\%\) confidence interval. Give your answers to two decimals \(\displaystyle?{<}{p}{<}?\)</span>

Answers (1)

2020-10-26
Step 1 The confidence interval for a sample proportion at \(\displaystyle{95}\%\) can be calculated using the formula, \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}\)</span>, where \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}\) is the sample proportion, N is the sample size and z* is the approprite value from the standard normal distributions for the desired confidence intervals. The z* value for \(\displaystyle{95}\%\) confidence interval is equal to 1.96 Step 2 The sample proportion is equal to 0.66 and the sample size is equal to 110 and the z* value is equal to 1.96 Substitute the values of \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}\), N and z* in the formula, \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}\)</span> to calculate the confidence interval of the sample proportion at \(\displaystyle{95}\%\) interval \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}-{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}{<}{p}{<}{w}{i}{d}{e}\hat{{{p}}}+{z}\cdot\sqrt{{{\frac{{{w}{i}{d}{e}\hat{{{p}}}{\left({1}-{w}{i}{d}{e}\hat{{{p}}}\right)}}}{{{N}}}}}}\)</span>
\(\displaystyle{0.66}-{\left({1.96}\sqrt{{{\frac{{{0.66}{\left({1}-{0.66}\right)}}}{{{110}}}}}}\right)}{<}{p}{<}{0.66}+{\left({1.96}\sqrt{{{\frac{{{0.66}{\left({1}-{0.66}\right)}}}{{{110}}}}}}\right)}\)</span>
\(\displaystyle{0.66}-{\left({1.96}\sqrt{{{0.00204}}}\right)}{<}{p}{<}{0.66}+{\left({1.96}\sqrt{{{0.00204}}}\right)}\)</span>
\(\displaystyle{0.66}-{\left({1.96}{\left({0.0451}\right)}{<}{p}{<}{0.66}+{\left({1.96}{\left({0.0451}\right)}\right.}\right.}\)</span>
\(\displaystyle{0.66}-{0.0884}{<}{p}{<}{0.66}+{0.0884}\)</span>
\(\displaystyle{0.5716}{<}{p}{<}{0.7484}\)</span> The confidence interval for the sample proportion at \(\displaystyle{95}\%\) is equal to (0.5716, 0.7484)
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